Calculate ΔHrxn for the following reaction: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) Use the following reactions and given ΔH values: 2Fe(s)+32O2(g)→Fe2O3(s),ΔH CO(g)+12O2(g)→CO2(g),ΔH==−824.2kJ−282.7kJ

Respuesta :

Answer : The value of [tex]\Delta H_{rxn}[/tex] is, -23.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is,

[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)[/tex]    [tex]\Delta H=?[/tex]

The intermediate balanced chemical reaction will be,

(1) [tex]2Fe(s)+\frac{3}{2}O_2(g)\rightarrow Fe_2O_3(s)[/tex]     [tex]\Delta H_1=-824.2kJ[/tex]

(2) [tex]CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)[/tex]    [tex]\Delta H_2=-282.7kJ[/tex]

We are reversing reaction 1 and multiplying reaction 2 by 3 and then adding all the equations, we get :

(1) [tex]Fe_2O_3(s)\rightarrow 2Fe(s)+\frac{3}{2}O_2(g)[/tex]     [tex]\Delta H_1=824.2kJ[/tex]

(2) [tex]3CO(g)+\frac{3}{2}O_2(g)\rightarrow 3CO_2(g)[/tex]    [tex]\Delta H_2=3\times (-282.7kJ)=-848.1kJ[/tex]

The expression for enthalpy of the reaction is:

[tex]\Delta H=\Delta H_1+\Delta H_2[/tex]

[tex]\Delta H=(824.2kJ)+(-848.1kJ)[/tex]

[tex]\Delta H=-23.9kJ[/tex]

Therefore, the value of [tex]\Delta H_{rxn}[/tex] is, -23.9 kJ